Today begins the second of three 2013 problems of the week. If you remember from last month, there are three different problems: easy, medium, and hard. The easy problem is at a level where you will be fine with just a good understanding of foundational mathematics (around a sixth grade level). The medium problem requires an understanding of Algebra I, but nothing beyond that. The hard problem requires Geometry, Algebra II, and depending on the rigor of the classes, some content from a Trigonometry or Precalculus class might be needed as well. To find your level, try looking at the categories of my blog posts. Basic posts correspond well with the easy problem, intermediate posts correspond with the medium, and advanced correspond with the hard.
Each of the problems is broken into five parts, and I give each part on the next day of the week. On Friday, you will have to solve for x, which is the final answer. The answer to the problem gets posted after a month. So, remember to save your results from each day to plug into the next day’s problem.
Last month’s easy problem had a lot of heavy arithmetic in it, which probably made it complicated to solve. However, this problem will be much less strenuous, and focuses more on the procedures than the arithmetic.
Take the following right triangle:
You are going to determine three things about this triangle. First, find the area, which will be denoted by the letter a.
a = ____
Now, find the perimeter, which will be p.
p = ____
Lastly, find the measure of the missing angle up top, which we’ll call m.
m = ____
Normally I would start with some triangle computations, but this week, I chose to do some function graphing on Monday, and then use triangles and trigonometry on Tuesday.
Take out some graph paper and draw in your x and y axes. Then, graph the following four functions, who we’ll call f1, f2, f3, and f4.
f1) y = 48 - 4/3x
f2) 10x - 24y - 360
f3) 3(x - 20) = -4y
f4) 5(y - 12) = 6(x + 15)
You won’t need to solve for any variables today, but keep the piece of paper that you graphed it on. This graph will be used in tomorrow’s problem.
Similar to the medium problem, today will be focused on graphing and tomorrow will incorporate the trigonometry.
Take out a piece of graph paper (separate from the medium problem if you are doing both) and graph the following equation:
16y2 = 3(48 - 3x2)
Tomorrow’s problem will be using this graph, so don’t lose it.